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gordonm888
gordonm888
Joined: Feb 18, 2015
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Thanks for this post from:
Gialmere
May 3rd, 2021 at 2:50:31 PM permalink
Late to this puzzle but:



There were 17 matches, because (14+12+8)/2 = 17

If Kenny only won 8 matches, then he must have lost every match, because a loser will play in alternate matches.

Kenny must have sat out match #1 and then played and lost in all even numbered matches, if he had played in odd number matches he would have played in 9 matches.

Thus Kenny lost the 4th match.

Abby won the last 7 matches but only played in 5 of the first ten, thus she lost her 5 matches (which were all odd numbered matches.)

Norman played in 14 matches, he can only have played in a maximum of 4 of the last seven matches, all of which Abby won. Thus he played in all 10 of the first 10 matches, winning them all.

So, in the 4th match, Norman beat Kenny.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
Ace2
Ace2
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May 3rd, 2021 at 8:31:50 PM permalink
Iím sure everyone is familiar with the St Petersburg paradox. If you get paid 2^x dollars the first time heads appears in a series of x coin flips, the EV is infinity.

Iíve noticed that the EV is finite for if you substitute any value less than e^.5 (about 1.65) for 2. Can someone explain why e^.5 is the upper limit for a finite EV?
Itís all about making that GTA
ThatDonGuy
ThatDonGuy
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May 4th, 2021 at 8:26:32 AM permalink
Quote: Ace2

Iím sure everyone is familiar with the St Petersburg paradox. If you get paid 2^x dollars the first time heads appears in a series of x coin flips, the EV is infinity.

Iíve noticed that the EV is finite for if you substitute any value less than e^.5 (about 1.65) for 2. Can someone explain why e^.5 is the upper limit for a finite EV?


It's not. The EV for any positive value n < 2 is n / (2 - n).

Let a be the number in place of 2 - i.e. you get a^n dollars if the first head appears on the nth toss.
The value = 1/2 x a + 1/4 + a^2 + 1/8 x a^3 + 1/16 x a^4 + ...
= a/2 + (a/2)^2 + (a/2)^3 + (a/2)^4 + ...
= a/2 (1 + a/2 + (a/2)^2 + a/2)^3 + ...)
If a >= 2, then a/2 >= 1, and the sum diverges.
If a < 2, then a/2 < 1, and the sum is 1 / (1 - a/2), so the value = (a/2) / (1 - a/2) = a / (2 - a).

Last edited by: ThatDonGuy on May 4, 2021
Ace2
Ace2
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May 4th, 2021 at 4:53:27 PM permalink
Quote: ThatDonGuy

It's not. The EV for any positive value n < 2 is n / (2 - n).


Let a be the number in place of 2 - i.e. you get a^n dollars if the first head appears on the nth toss.
The value = 1/2 x a + 1/4 + a^2 + 1/8 x a^3 + 1/16 x a^4 + ...
= a/2 + (a/2)^2 + (a/2)^3 + (a/2)^4 + ...
= a/2 (1 + a/2 + (a/2)^2 + a/2)^3 + ...)
If a >= 2, then a/2 >= 1, and the sum diverges.
If a < 2, then a/2 < 1, and the sum is 1 / (1 - a/2), so the value = (a/2) / (1 - a/2) = a / (2 - a).

I used the integral from zero to infinity of .5 * e^(-x/2) * a^x. For a values under 1.2, it comes out quite close to a / (2 - a). But apparently my answer is for the exponential distribution only and does not carry over to the binomial (it usually does). I thought this would work...maybe the Wizard can explain why it doesnít?
Itís all about making that GTA
Gialmere
Gialmere
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May 5th, 2021 at 5:39:44 PM permalink
Here's an easy one similar to the pickleball puzzle...



Adam and Eve play rock-paper-scissors 10 times. You know that:

Adam uses rock three times, scissors six times, and paper once.
Eve uses rock twice, scissors four times, and paper four times.
There are no ties in all 10 games.
The order of games is unknown.


Who wins? By how much?
Have you tried 22 tonight? I said 22.
ThatDonGuy
ThatDonGuy
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Thanks for this post from:
Gialmere
May 5th, 2021 at 6:40:14 PM permalink
Quote: Gialmere

Here's an easy one similar to the pickleball puzzle...

Adam and Eve play rock-paper-scissors 10 times. You know that:

Adam uses rock three times, scissors six times, and paper once.
Eve uses rock twice, scissors four times, and paper four times.
There are no ties in all 10 games.
The order of games is unknown.


Who wins? By how much?



Eve uses scissors four times. Since there are no ties, they have to be the four times Adam did not use scissors.

The breakdown is:
Adam uses scissors, Eve uses rock: 2 times (Eve wins)
Adam uses scissors, Eve uses paper: 4 times (Adam wins)
Adam uses rock, Eve uses scissors: 3 times (Adam wins)
Adam uses paper, Eve uses scissors: 1 time (Eve wins)

Adam wins 7 games; Eve wins 3

Gialmere
Gialmere
Joined: Nov 26, 2018
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May 5th, 2021 at 9:27:15 PM permalink
Quote: ThatDonGuy



Eve uses scissors four times. Since there are no ties, they have to be the four times Adam did not use scissors.

The breakdown is:
Adam uses scissors, Eve uses rock: 2 times (Eve wins)
Adam uses scissors, Eve uses paper: 4 times (Adam wins)
Adam uses rock, Eve uses scissors: 3 times (Adam wins)
Adam uses paper, Eve uses scissors: 1 time (Eve wins)

Adam wins 7 games; Eve wins 3


Correct!
-----------------------------------------

"Mom, how did humans come to exist?"

"Well, you see, God created Adam and Eve..."

"But dad said we came from apes."

"He was talking about his side of the family, I'm telling you about mine."
Have you tried 22 tonight? I said 22.
DogHand
DogHand
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May 6th, 2021 at 9:43:08 AM permalink


Speaking of Adam and Eve... why does Eve have a navel?

Dog Hand

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